Problem: Solve for $x$ and $y$ using substitution. ${5x+3y = 12}$ ${x = 5y+8}$
Since $x$ has already been solved for, substitute $5y+8$ for $x$ in the first equation. ${5}{(5y+8)}{+ 3y = 12}$ Simplify and solve for $y$ $25y+40 + 3y = 12$ $28y+40 = 12$ $28y+40{-40} = 12{-40}$ $28y = -28$ $\dfrac{28y}{{28}} = \dfrac{-28}{{28}}$ ${y = -1}$ Now that you know ${y = -1}$ , plug it back into $\thinspace {x = 5y+8}\thinspace$ to find $x$ ${x = 5}{(-1)}{ + 8}$ $x = -5 + 8$ ${x = 3}$ You can also plug ${y = -1}$ into $\thinspace {5x+3y = 12}\thinspace$ and get the same answer for $x$ : ${5x + 3}{(-1)}{= 12}$ ${x = 3}$